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Find S10 for the series 2 + 7 + 12 + 17 +...
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\[\begin{align*}S_{10}&=\sum_{n=1}^{10}(2+5(n-1))\\ &=2\sum_{n=1}^{10}1+5\left(\sum_{n=1}^{10}n-\sum_{n=1}^{10}1\right) \end{align*}\] Now use the following formulas: \[\sum_{n=1}^{k}1=k\\ \sum_{n=1}^{k}n=\frac{k(k+1)}{2}\] (In this case, \(k=10\).)
sum = n/2(2a + (n - 1)d n = 10 d = 5 a = 2 lets sub it in... s10 = 10/2 * (2*2 + (10 - 1)* 5 s10 = 5 * (4 + 9) * 5 s10 = 5 (13 * 5) s10 = 5(26) s10 = 130
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